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secondary 3 | A Maths
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Date Posted: 4 years ago
Views: 626
(i)
use t=0,
principal amt = 50 000
(ii)
find p first,
use t=1,
A = 50 000(1.5)ᵖ = 80 000
=> (1.5)ᵖ = 1.6
21 mths = 1.75 = t
A = 50 000 (1.5)ᵖᵗ = 50 000 x (1.6)¹⁷⁵
= 113 809.882
(iii)
A = 50 000(1.5)ᵖᵗ > 1000 000
=> 50 000 x (1.6)¹⁷⁵ᵗ > 1000 000
=> (1.6)¹⁷⁵ᵗ > 20
=> 1.75t log (1.6) > log 20
=> t > 3.642 yr
=> t > 3 yr 7.7 mths
min. 3 yrs 8 mths or 44mths
use t=0,
principal amt = 50 000
(ii)
find p first,
use t=1,
A = 50 000(1.5)ᵖ = 80 000
=> (1.5)ᵖ = 1.6
21 mths = 1.75 = t
A = 50 000 (1.5)ᵖᵗ = 50 000 x (1.6)¹⁷⁵
= 113 809.882
(iii)
A = 50 000(1.5)ᵖᵗ > 1000 000
=> 50 000 x (1.6)¹⁷⁵ᵗ > 1000 000
=> (1.6)¹⁷⁵ᵗ > 20
=> 1.75t log (1.6) > log 20
=> t > 3.642 yr
=> t > 3 yr 7.7 mths
min. 3 yrs 8 mths or 44mths
For part (iii), it's 1.6^t, not 1.6^1.75t. The 1.75 was just a value for usage in part ii.
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Something like this would be good. I have deviated from part iii's requirement because a rounding down of the value (to the question's desire) does not exactly make sense here.
Date Posted:
4 years ago
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